If \log_4 x + \log_2 y = \frac{5}{2} and 3^x \cdot 9^{-y} = 81, what is the value of x + y… — JAMB UTME

If $\log_4 x + \log_2 y = \frac{5}{2}$ and $3^x \cdot 9^{-y} = 81$ , what is the value of $x + y$ ?

The value of

x + y

10

Converting

\log_4 x

to base

2

gives

\frac{1}{2}\log_2 x

, leadings to

x y^2 = 32

. Simplifying

3^x \cdot 9^{-y} = 81

gives

x - 2y = 4

. Solving these simultaneously gives

x = 8

and

y = 2

, whose sum is

10

Simplify the logarithmic equation using the change of base formula.

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x

. Substituting this gives

\frac{1}{2}\log_2 x + \log_2 y = \frac{5}{2} \implies \log_2 x + 2\log_2 y = 5

Converting logarithms to a common base of

2

allows combining terms.

Combine logarithmic terms and express as an algebraic relation.

\log_2(x y^2) = 5 \implies x y^2 = 2^5 = 32

Applying the logarithmic laws

k\log_b a = \log_b(a^k)

and

\log_b M + \log_b N = \log_b(MN)

Simplify the exponential equation.

3^x \cdot (3^2)^{-y} = 3^4 \implies 3^{x - 2y} = 3^4 \implies x - 2y = 4

Expressing both sides with base

3

allows equating exponents.

Solve the system of equations for

x

and

y

From

x = 4 + 2y

, substitute into

x y^2 = 32

(4 + 2y)y^2 = 32 \implies 2y^3 + 4y^2 - 32 = 0 \implies y^3 + 2y^2 - 16 = 0

. Testing positive factors yields

y = 2

. Then

x = 4 + 2(2) = 8

Simultaneous substitution yields the unique real positive solutions

x = 8

and

y = 2

Calculate

x + y

x + y = 8 + 2 = 10

Adding the computed values of

x

and

y

gives the required sum.

Simultaneous Exponential and Logarithmic Systems

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